\(a,\left(3x+1\right)^2=x^2\)

=>  \(\left(3x+1\right)^2-x^2=0\)

=>  \(\left(3x+1-x\right)\left(3x+1+x\right)=0\)

=>  \(\left(2x+1\right)\left(4x+1\right)=0\)

=> \(\orbr{\begin{cases}2x+1=0\\4x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{1}{4}\end{cases}}}\)

\(b,x^2-8x+16=0\)

=>  \(\left(x-4\right)^2=0\)

=>  \(x-4=0\)

=>  \(x=4\)

\(a,73^2-13^2-10^2+20\cdot13=\left(73-13-10\right)^2+20\cdot13=50^2+260=2500+260=2760\)

Ta có :

\(73^2-13^2-10^2+20.13=\left(73+13\right)\left(73-13\right)-13.13+20.13\)

\(=86.60-13\left(13-20\right)\)

\(=5160-\left(-91\right)=5160+91=...\)

\(a,\left(x^2+4\right)^2-x^2=\left(x^2+4-x\right)\left(x^2+4+x\right)\)

\(b,\left(3x-2y\right)^2-\left(2x-3y\right)^2=\left(3x-2y-2x+3y\right)\left(3x-2y+2x-3y\right)\)

\(=\left(x+y\right)\left(5x-5y\right)=5\left(x+y\right)\left(x-y\right)\)

\(c,\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)

\(d,8x^3+1=\left(2x\right)^3+1^3=\left(2x+1\right)\left(4x^2-2x+1\right)\)

Mình cck cho 5 bạn đầu tiên nha

quy đồng làm mẫu số hai bên giống nhau

quy đồng hai số bên cho giống nhau

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hot tốt

\(\frac{7}{4}ha-\frac{1}{2}ha=\frac{5}{2}ha=\frac{5}{2}\cdot10dam=25dam\)

xin lỗi là \(dam^2\)     nha

bằng 43/7 nha